Each of the 200 students in a class enrolled in at least one of the th

Question

Each of the 200 students in a class enrolled in at least one of the three subjects: Physics, Chemistry and Mathematics. If 30 participated in, Physics, 150 participated in Chemistry and 100 participated in Mathematics and if “x” participated in all the three activities, then what could be the maximum value of x?

A. 10
B. 20
C. 30
D. 40
E. 50

A. 10
B. 20
C. 30
D. 40
E. 50

doomedcat · Answer

max value of x can be when all of the students in least # count of the subjects enrol for all 3 subjects
in this case its physics 30 ans C

30 is the limiting factor as any more than 30 would not suffice all three enrolment

rsrighosh · Answer

VeritasKarishma

I was going through the question
https://gmatclub.com/forum/there-are-10 ... 58096.html

The question is:-

There are 101 students in a school. The students are split into 3 teams. Team A contains 70 students, team B contains 75 students and team C contains 80 students. What is maximum number of students who can be present in all three teams?

A. 70
B. 65
C. 63
D. 62
E. 61[/
color]

[color=#a0410d]Answer:-

You have to find maximum A&B&C, meaning you have to minimise "exactly 2" i.e only A&B + only B&C + only C&A

Total = A + B + C - A&B - B&C - C&A + A&B&C

exactly 2 = A&B + B&C + C&A - 3(A&B&C)
=> A&B + B&C + C&A = exactly 2 + 3(A&B&C)

Hence the first equation can be written as

Total = A + B + C - A&B - B&C - C&A + A&B&C
=> Total = A + B + C - exactly 2 - 3(A&B&C) + A&B&C
=> Total = A + B + C - exactly 2 - 2(A&B&C)

This is how you get the formula

Now replacing values
101 = 70 + 75 + 80 - exactly 2 - 2(A&B&C)
=> 124 = exactly 2 + 2(A&B&C)

To maximise A&B&C, replace exactly 2 as 0

124 = 0 + 2(A&B&C)
=> (A&B&C) = 62

a)How is this different from the current question?

b) Isn't 30 the minimum value?

c) if i work this problem with Total = A + B + C - exactly 2 - 2(A&B&C), the answer is 40. shouldn't this be answer to maximum?

Kindly help. Looking forward to your reply

flapjack29 · Answer

It’s a trick here. If you see, max amount of people taking 1 class is 30. Therefore even though x =40 we are limited to the parameters of this problem.

KarishmaB · Answer

The questions are very similar with similar parameters. The only thing different is the numbers given. But because of the numbers given, we are using different methods on the two questions.

Here, one set is very very small compared to the other two which are fairly large. 100 and 150 will certainly have an overlap of 50 because there are 200 students total. So if you make the set 30 overlap on the 50, you get the overlap of all 3 as 30 (which is the max possible because only 30 take physics). The other 20 of the 50 will have 2 subjects - Chem and Math. You cannot have exactly 2 = 0 here.

In the linked question, the total was 110 and all sets were close to each other. A lot more overlap is possible of all 3 there and hence we were able to make exactly 2 = 0 in that case.

If you want one consistent method, when maximising the overlap of all three, put all circles one inside the other. Then if none needs to be 0 but is not, see from where will you pull out values. 
Here, for example, put the 150 circle, then 100 circle inside it and then 30 circle inside the 100 circle (so 70 are outside). This gives you an overlap of 30 for all 3. 
But none = 50 here so ensure that you pull some values out. You can pull 50 values from the 70 out to cover the 50. 
Your overlap of all 3 stays at 30.

P.S. - I am glad you are creating these links and thinking through on various tweaks utilised in different questions.

bumpbot · Answer

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Each of the 200 students in a class enrolled in at least one of the th

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